Áp dụng BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\):

\(VT=\sqrt{\frac{x^2+\left(2y\right)^2}{2}}+\sqrt{\frac{\left(\frac{x}{2}-y\right)^2+3\left(\frac{x}{2}+y\right)^2}{3}}\)

\(VT\ge\sqrt{\frac{\left(x+2y\right)^2}{4}}+\sqrt{\frac{3\left(\frac{x}{2}+y\right)^2}{3}}\)

\(VT\ge\left|\frac{x+2y}{2}\right|+\left|\frac{x+2y}{2}\right|=\left|x+2y\right|\ge x+2y\) (đpcm)

Dấu "=" xảy ra khi \(x=2y\ge0\)

\(x^2+2y^2-2xy+4y+5=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)+1\)

\(=\left(x-y\right)^2+\left(y+2\right)^2+1\ge1>0\)

=>đpcm

ta có 4x² + 4y² + 6x + 3 ≥ 4xy

<=> (x² - 4xy + 4y²) + 3(x² + 2x + 1) ≥ 0

<=> (x - 2y)² + 3(x +1)² ≥ 0 (luôn đúng với mọi x,y

vậy với mọi x,y ta có 4x² + 4y² +6x + 3 ≥ 4xy

Theo bài ra :4x²+4y²+6x+3≥4xy

⇔4x²+4y²+6x +3 -4xy≥0 ⇔ [x²-4xy+(2y)²] +3x²+6x+3≥0

⇔(x-2y)²+3(x²+2x+1)≥0  ⇔ (x-2y)² +3(x+1)²  ≥0 ,∀ x,y

vậy 4x²+4y²+6x+3≥4xy

$4x^2+4y^2+4xy>6y-4\left(1\right)$

$\iff 4x^2+4y^2+4xy-6y+4>0\left(2\right)$

$\iff 4x^2+4xy+y^2+3y^2-6y+3+1>0$

$\iff {(2x+y)}^2+3(y^2-2y+1)+1>0$

$\iff {(2x+y)}^2+3{(y-1)}^2+1>0$

$+){(2x+y)}^2\ge 0$

$3{(y-1)}^2\ge 0$

$\to {(2x+y)}^2+3{(y-1)}^2\ge 0$

$\to {(2x+y)}^2+3{(y-1)}^2+1\ge 1>0$

BĐT(2) luôn đúng

$\to$ BĐT(1) luôn đúng

Vậy $4x^2+4y^2+4xy>6y-4$

\(x^2+2y^2-2xy+2x-4y+2=0\)

\(\Rightarrow x^2-2xy+y^2+2\left(x-y\right)+1+y^2-2y+1=0\)

\(\Rightarrow\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-1\right)^2=0\)

\(\Rightarrow\left(x-y+1\right)^2+\left(y-1\right)^2=0\)

=>................

viết đề khó hiểu quá

\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)

\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)

ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)

Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)

T i c k cho mình 1 cái nha mới bị trừ 50 đ

Đặt \(f\left(x\right)=25x^2+25y^2+9x^2+16y^2+144-72x-96y+24xy-72\)

\(=34x^2+41y^2-72x-96y+24xy+72\)

\(=34x^2+2\left(12y-36\right)x+41y^2-96y+72\)

\(a=34>0\)

\(\Delta'=\left(12y-36\right)^2-34\left(41y^2-96y+72\right)\)

\(=-1250y^2+2400y-1152=-2\left(25y-24\right)^2\le0;\forall y\)

\(\Rightarrow f\left(x\right)\ge0;\forall x;y\)